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mr_godi [17]
4 years ago
14

A ball is dropped and falls with an acceleration of 9.8 m/s2 downward. It hits the ground with a velocity of 49 m/s downward. Ho

w long did it take the ball to fall to the ground?
Physics
1 answer:
il63 [147K]4 years ago
6 0
Hey!

NOTE-:

u= initial velocity
v= final velocity
g= acceleration due to gravity
t= time

u= 0
v= 49 m/s
t=?
g= 9.8 m/s^2

Using first equation of motion -

v-u=at
49-0= 9.8×t
49 = 9.8t
49/9.8= t
t= 5 second


Hope it helps...!!!
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A 1 036-kg satellite orbits the Earth at a constant altitude of 98-km. (a) How much energy must be added to the system to move t
Veronika [31]

Answer:

a) The Energy added should be 484.438 MJ

b) The  Kinetic Energy change is -484.438 MJ

c) The Potential Energy change is 968.907 MJ

Explanation:

Let 'm' be the mass of the satellite , 'M'(6×10^{24} be the mass of earth , 'R'(6400 Km) be the radius of the earth , 'h' be the altitude of the satellite and 'G' (6.67×10^{-11} N/m) be the universal constant of gravitation.

We know that the orbital velocity(v) for a satellite -

v=\sqrt{\frac{Gmm}{R+h} }         [(R+h) is the distance of the satellite   from the center of the earth ]

Total Energy(E) = Kinetic Energy(KE) + Potential Energy(PE)

For initial conditions ,

h = h_{i} = 98 km = 98000 m

∴Initial Energy (E_{i})  = \frac{1}{2}mv^{2} + \frac{-GMm}{(R+h_{i} )}

Substituting v=\sqrt{\frac{GMm}{R+h_{i} } } in the above equation and simplifying we get,

E_{i} = \frac{-GMm}{2(R+h_{i}) }

Similarly for final condition,

h=h_{f} = 198km = 198000 m

∴Final Energy(E_{f}) = \frac{-GMm}{2(R+h_{f}) }

a) The energy that should be added should be the difference in the energy of initial and final states -

∴ ΔE = E_{f} - E_{i}

        = \frac{GMm}{2}(\frac{1}{R+h_{i} } - \frac{1}{R+h_{f} })

Substituting ,

M = 6 × 10^{24} kg

m = 1036 kg

G = 6.67 × 10^{-11}

R = 6400000 m

h_{i} = 98000 m

h_{f} = 198000 m

We get ,

ΔE = 484.438 MJ

b) Change in Kinetic Energy (ΔKE) = \frac{1}{2}m[v_{f} ^{2} - v_{i} ^{2}]

                                                          = \frac{GMm}{2}[\frac{1} {R+h_{f} } - \frac{1} {R+h_{i} }]

                                                          = -ΔE                                                            

                                                          = - 484.438 MJ

c)  Change in Potential Energy (ΔPE) = GMm[\frac{1}{R+h_{i} } - \frac{1}{R+h_{f} }]

                                                             = 2ΔE

                                                             = 968.907 MJ

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How do you use the lever (simple machines)
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Answer:

You pull it and it activates something like a light.That's what you do with a lever.

Explanation:

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How can you find the net force if two forces act in opposite direction?
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Subtract one from the to find the magnitude. Whether it's positive or negative will determine the direction.
5 0
4 years ago
Read 2 more answers
The force of gravity pulls down on your school with a total force of 400,000 newtons. The force of gravity pulling down on your
gizmo_the_mogwai [7]

Answer: a Had twice as much mass

Explanation:

The data that we have is:

"The force of gravity pulls down on your school with a total force of 400,000 newtons. "

First, remember that, by the second Newton's law that:

F = a*m

F = force

a = acceleration

m = mass

In the case of the gravitational force, the gravitational acceleration is a constant: a = 9.8m/s^2

Then, if we want to have twice as much force the only thing that we can change in the equation is the mass:

Then if the initial force is written as:

F = a*m

twice as much that force is written as:

2*F = a*x

x is a variable that represents the new mass.

We know that F = a*m

2*F = 2*a*m

2*a*m  = a*x

2*m = x

Then, if we want to have twice as much force, we should have twice as much mass.

7 0
3 years ago
A 25 kg child stands 2.5 m from the center of a frictionless merry‐go‐round, which has a 200 kg*m^2 moment of inertia and is spi
vodka [1.7K]

Answer:

Explanation:

a ) Time period  T = 2 s

Angular velocity ω = 2π / T

=  2π / 2 = 3.14 rad /s

Initial moment of inertia I₁ = 200 + mr²

= 200 + 25 x 2.5²

=356.25

Final moment of inertia

I₂ = 200 + 25 X 1.5 X 1.5

= 256.25

b ) We apply law of conservation of momentum

I₁ X ω₁ =  I₂ X ω₂

ω₂ = I₁ X ω₁ / I₂

Putting the values

w_2=\frac{356.25\times3.14}{256.25}

ω₂ = 4.365 rad s⁻¹

c ) Increase in rotational kinetic energy

=1/2 I₂ X ω₂² -  1/2 I₁ X ω₁²

.5 X 256.25 X 4.365² - .5 X 356.25 X 3.14²

= 684.95 J

This energy comes from work done against the centripetal pseudo -force.

7 0
3 years ago
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